Product of cosines: $\prod_{r=1}^{7} \cos \left(\frac{r\pi}{15}\right)$

Evaluate

$$\prod_{r=1}^{7} \cos \left({\dfrac{r\pi}{15}}\right)$$

I tried trigonometric identities of product of cosines, i.e, $$\cos\text{A}\cdot\cos\text{B} = \dfrac{1}{2}[ \cos(A+B)+\cos(A-B)]$$

but I couldn't find the product.

Any help will be appreciated.
Thanks.

• Hint: Consider the square of the product. And consider the leading coefficient of the Chebyshev Polynomial $\cos(15x)$ when written in terms of $\cos(x)$. Commented Jul 6, 2015 at 14:59
• @GohP.iHan Can you please elaborate your method in a solution, it looks interesting. Thanks. Commented Jul 6, 2015 at 15:20
• If you square it, then it is of the form of $-\cos(\pi/15) \cos(2\pi/15)\ldots\cos(14\pi/15)$. Now consider $\cos(15x) = 0$ which has all the 14 roots. Writing it as the Chebyshev polynomial, $0=\cos(15x) = 2^{14}\cos^{15}(x) -\ldots -1$. By Vieta's formula, the product is simply $1/2^{14}$, take its square root, and voilà! Commented Jul 6, 2015 at 15:57
• @GohP.iHan Thank you so much! It is enlightening to know various approaches to the same problem :) Commented Jul 6, 2015 at 18:24

Let $\displaystyle\text{C}=\prod_{r=1}^{7}\cos{\left(\dfrac{r\pi}{15}\right)}$

and

$\displaystyle\text{S}=\prod_{r=1}^{7}\sin{\left(\dfrac{r\pi}{15}\right)}$

Now,

$\text{C}\cdot\text{S}=\left(\sin{\dfrac{\pi}{15}} \cdot \cos{\dfrac{\pi}{15}}\right) \cdot \left(\sin{\dfrac{2\pi}{15}}\cdot\cos{\dfrac{2\pi}{15}}\right)\cdot \ldots \cdot\left(\sin{\dfrac{7\pi}{15}} \cdot \cos{\dfrac{7\pi}{15}}\right)$

$\implies \text{C}\cdot\text{S}= \dfrac{1}{2^7} \left(2\sin{\dfrac{\pi}{15}} \cdot \cos{\dfrac{\pi}{15}}\right) \cdot \left(2\sin{\dfrac{2\pi}{15}}\cdot\cos{\dfrac{2\pi}{15}}\right)\cdot \ldots \cdot\left(2\sin{\dfrac{7\pi}{15}} \cdot \cos{\dfrac{7\pi}{15}}\right)$

$\implies \text{C}\cdot\text{S}= \dfrac{1}{2^7} \ \sin{\dfrac{2\pi}{15}}\cdot\sin{\dfrac{4\pi}{15}}\cdot \ldots \cdot\sin{\dfrac{14\pi}{15}}$

$\{\because \sin(2x) = 2\sin (x) \cos (x) \}$

$\implies \text{C}\cdot\text{S}= \dfrac{1}{2^7} \ \sin{\dfrac{\pi}{15}}\cdot\sin{\dfrac{2\pi}{15}} \cdot \ldots \cdot \sin{\dfrac{7\pi}{15}} \\\\ \{\because \sin(\pi-x)=\sin(x)\}$

$\implies \text{C}\cdot\text{S}= \dfrac{1}{2^7} \cdot \text{S}$

since $\text{S} \neq 0$,

$\therefore \boxed{\text{C}=\dfrac{1}{2^7}}$

• +1 Looks like this can be generalized to $$\prod_{r=1}^n \cos\frac{r\pi}{2n+1} = \frac{1}{2^n}$$. Commented Jul 6, 2015 at 14:29
• (+1) very nice approach, it reminds the Gauss lemma for quadratic reciprocity - in particular, the proof that $$\left(\frac{2}{p}\right)=(-1)^{\frac{p^2-1}{8}}.$$ Commented Jul 6, 2015 at 14:57
• @Ishu Using math.stackexchange.com/questions/159214/… then: $$\frac{2n}{2^{2n}} = \prod_{r=1}^{2n} \sin \frac{r\pi}{2n+1} = \left(\prod_{r=1}^{n} \sin \frac{r\pi}{2n+1}\right)\left(\prod_{r=n+1}^{2n} \sin \frac{r\pi}{2n+1}\right) = \left(\prod_{r=1}^{n} \sin \frac{r\pi}{2n+1}\right)\left(\prod_{r=1}^{n} \sin \frac{(n+r)\pi}{2n+1}\right)$$ And since $\sin \dfrac{(n+r)\pi}{2n+1} = \sin \dfrac{(n+1-r)\pi}{2n+1}$ then $$\dfrac{2n}{2^{2n}} = \left(\prod_{r=1}^{n} \sin \frac{r\pi}{2n+1}\right)^2$$ Therefore the product of sines is $\dfrac{\sqrt{2n}}{2^n}$. Commented Jul 6, 2015 at 15:22
• EDIT: It should be $\dfrac{2n+1}{2^{2n}} = \ldots$ and the product of sines is therefore $\dfrac{\sqrt{2n+1}}{2^n}$. Commented Jul 6, 2015 at 19:18
• +1 This answer helped me to solve exercise 5.32 of Ireland and Rosen's number theory book. I find it quite interesting that by replacing $\cos\frac{r\pi}{2n+1}$ with $\cos\frac{2\pi r}{2n+1}$ and requiring that $2n+1$ is prime, suddenly we get a formula for the quadratic character of $2$ mod that prime! Commented Jul 1, 2022 at 19:50

I think it's worth noting the product is also evaluable just remembering, besides the well known $\displaystyle \cos\frac{\pi}{3}=\frac{1}{2},$ the somewhat nice $$\displaystyle\cos\frac{\pi}{5}=\frac{1+\sqrt{5}}{4}=\frac{\phi}{2}$$ (where $\phi$ is the golden section) and iterating the sum/difference formula for the cosine and the product formula you mention. Your product is, once we rearrange factors and simplify fractions, equal to $${\color\red{\cos\frac{\pi}{3}\cdot\cos\frac{\pi}{5}}}\cdot\color\orange{\cos\frac{2\pi}{5}} \cdot\color\navy{\cos\frac{\pi}{15}\cdot\cos\frac{4\pi}{15}}\cdot\color\green{\cos\frac{2\pi}{15}\cdot\cos\frac{7\pi}{15}} \\ ={\color\red{\frac{\phi}{4}}}\color\orange{\left(2\cos^2\frac{\pi}{5}-1\right)}\color\navy{\frac{1}{2}\left(\cos\frac{\pi}{3}+\cos\left(-\frac{\pi}{5}\right)\right)}\color\green{\frac{1}{2}\left(\cos\left(-\frac{\pi}{3}\right)+\cos\frac{3\pi}{5}\right)} \\ = \frac{\phi}{16}\left(\frac{\phi^2}{2}-1\right)\frac{\phi+1}{2}\left(\frac{1}{2}+\cos\frac{\pi}{5}\cdot\left(2\cos^2\frac{\pi}{5}-1\right)-2\cos\frac{\pi}{5}\left(1-\cos^2\frac{\pi}{5}\right)\right)\\=\frac{\phi(\phi^2-1)}{64}\left(\frac{1}{2}+\frac{\phi(\phi^2-2)}{4}-\frac{\phi(4-\phi^2)}{4}\right)\\=\frac{\phi^2}{64}\left(\frac{1}{2}+\frac{\phi(\phi-1)-\phi(3-\phi)}{4}\right)\\=\frac{\phi+1}{64}\left(\frac{1}{2}+\frac{2-2\phi}{4}\right)=\frac{\phi+1}{128}-\frac{\phi^2-1}{128}=\frac{\phi+1}{128}-\frac{\phi}{128}=\frac{1}{128}.$$

Since an elegant solution has already been provided, I will go for an overkill.

From the Fourier cosine series of $$\log\cos x$$ we have: $$\log\cos x = -\log 2-\sum_{n\geq 1}^{+\infty}\frac{(-1)^n\cos(2n x)}{n}\tag{1}$$ but for any $$n\geq 1$$ we have: $$15\nmid n\rightarrow\sum_{k=1}^{7}\cos\left(\frac{2n k \pi}{15}\right) = -\frac{1}{2},\quad 15\mid n\rightarrow\sum_{k=1}^{7}\cos\left(\frac{2n k \pi}{15}\right) = 7\tag{2}$$ so: $$\sum_{k=1}^{7}\log\cos\frac{k\pi}{15} = -7\log 2+\frac{1}{2}\sum_{n\geq 1}\frac{(-1)^n}{n}-\frac{15}{2}\sum_{n\geq 1}\frac{(-1)^n}{15n}=-7\log 2\tag{3}$$ and by exponentiating the previous line:

$$\prod_{k=1}^{7}\cos\left(\frac{\pi k}{15}\right) = \color{red}{\frac{1}{2^7}}.\tag{4}$$

• Can you tell me how did you arrive at $(2)$ ? Commented Jul 6, 2015 at 15:23
• @Samurai: write $\cos z$ as $\text{Re}(e^{iz})$. Then the sum becomes a trivial sum if $15\mid n$, and half a sum of $15$-th roots of unity if $15\nmid n$. Commented Jul 6, 2015 at 15:25

Note: Here's another variation inspired by an answer to this question.

We consider the roots of unity $e^{\frac{2\pi i k}{15}}, 0\leq k < 15$ of the polynomial

$$p(z)=z^{15}-1=\prod_{k=0}^{14}(z-e^{\frac{2\pi i k}{15}})$$

We obtain

\begin{align*} -p(-z)=z^{15}+1&=\prod_{k=0}^{14}(z+e^{\frac{2\pi i k}{15}})\\ &=(z+1)\prod_{k=1}^{7}\left[(z+e^{\frac{2\pi i k}{15}})(z+e^{-\frac{2\pi i k}{15}})\right]\\ \end{align*}

Evaluating the polynomial $-p(-z)$ at $z=1$ gives

\begin{align*} 1&=\prod_{k=1}^{7}\left[(1+e^{\frac{2\pi i k}{15}})(1+e^{-\frac{2\pi i k}{15}})\right]\\ &=\prod_{k=1}^{7}\left[(e^{-\frac{\pi i k}{15}}+e^{\frac{\pi i k}{15}})e^{\frac{\pi i k}{15}}(e^{\frac{\pi i k}{15}}+e^{-\frac{\pi i k}{15}})e^{-\frac{\pi i k}{15}}\right]\\ &=\prod_{k=1}^{7}(e^{\frac{\pi i k}{15}}+e^{-\frac{\pi i k}{15}})^2\tag{1}\\ &=\prod_{k=1}^{7}\left(2\cos\left(\frac{k \pi}{15}\right)\right)^2\tag{2}\\ \end{align*}

In (1) we use the formula $\cos(z)=\frac{1}{2}\left(e^{iz}+e^{-iz}\right)$.

We conclude from (2) \begin{align*} \prod_{k=1}^{7}\cos\left(\frac{k\pi}{15}\right)=\frac{1}{2^7} \end{align*}

Note: Writing $-p(-z)$ as

\begin{align*} -p(-z)=\prod_{k=1}^{7}\left[z^2+\left(e^{\frac{2\pi i k}{15}}+e^{-\frac{2\pi i k}{15}}\right)z+1\right]\\ \end{align*}

and evaluating the polynomial $-p(-z)$ at $z=i$ we obtain the related formula

\begin{align*} \prod_{k=1}^{7}\cos\left(\frac{2k\pi}{15}\right)=\frac{1}{2^7} \end{align*}

Doubling the argument does not change the value of the product.

I like the answer by @Steven Gregory because of the way the dominos fall and it seems the only one presented that a precalculus student could hope to find. Using the reflection and multiplication formulas for the gamma function a fairly compact proof is possible. \begin{align}\prod_{r=1}^7\cos\frac{r\pi}{15} & = \prod_{r=1}^7\sin\left(\frac{\pi}2-\frac{r\pi}{15}\right) = \prod_{r=0}^6\sin\left(\frac{\pi}2-\frac{7\pi}{15}+\frac{r\pi}{15}\right) \\ & = \prod_{r=0}^6\sin\left(\frac{\pi}{30}+\frac{r\pi}{15}\right) = \prod_{r=0}^6\frac{\pi}{\Gamma\left(\frac1{30}+\frac{r}{15}\right)\Gamma\left(\frac{29}{30}-\frac{r}{15}\right)} & \tag{1} \\ & = \frac{\pi^7}{\prod_{r=0}^6\Gamma\left(\frac1{30}+\frac{r}{15}\right)\prod_{s=8}^{14}\Gamma\left(\frac1{30}+\frac{s}{15}\right)} = \frac{\pi^7\,\Gamma\left(\frac1{30}+\frac7{15}\right)}{\prod_{r=0}^{14}\Gamma\left(\frac1{30}+\frac{r}{15}\right)} & \tag{2} \\ & = \frac{\pi^7\,\Gamma\left(\frac12\right)}{(2\pi)^{(15-1)/2}15^{\frac12-15\left(\frac1{30}\right)}\Gamma\left(15\left(\frac1{30}\right)\right)} & \tag{3} \\ & = \frac1{2^7} \end{align} $(1)$ Using the reflection formula for the gamma function: $\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin(\pi x)}$.
$(2)$ Reordering one product and multiplying and dividing by $\Gamma\left(\frac1{30}+\frac7{15}\right)$.
$(3)$ Using the multiplication formula for the gamma function: $$\prod_{k=0}^{n-1}\Gamma\left(x+\frac{k}n\right)=(2\pi)^{(n-1)/2}n^{\frac12-nx}\Gamma(nx)$$

• (+1) Nice. But can you explain how do you find the gamma products? IMO the accepted answer is the most elegant for anyone, precalc or calculus student. Commented Mar 21, 2016 at 4:04
• Hope my edits clarified the situation a bit. This is kind of a juicy product in that at first glance it just looks outrageous, but there are actually many disparate methods of attack, all of which seem to work. Commented Mar 21, 2016 at 6:29
• The proofs of the multiplication formula that I know rely on a calculation equivalent to the obvious generalization of the one OP asks about, so this seems a bit circular to me. Do you know a proof of hte multiplication formula that doesn't work that way? Commented Jul 21, 2020 at 18:50

Using de Moivre's formula,

$$\cos(2n+1)x+i\sin(2n+1)x=(\cos x+i\sin x)^{2n+1}=\sum_{r=0}\binom{2n+1}r(\cos x)^{2n+1-r}(i\sin x)^r$$

Equating the real parts, $$\cos(2n+1)x$$

$$=(\cos x)^{2n+1}-\binom{2n+1}2(\cos x)^{2n+1-2}(\sin x)^2+\binom{2n+1}r(\cos x)^{2n+1-4}(\sin x)^4+\cdots+\binom{2n+1}{2n-2}(-1)^{n-1}(\cos x)^3(\sin x)^{2n-2}+(-1)^n\cos x\sin^{2n}x$$

$$=(\cos x)^{2n+1}-\binom{2n+1}2(\cos x)^{2n+1-2}(1-\cos^2x)+\binom{2n+1}r(\cos x)^{2n+1-4}(1-\cos^2x)^2+\cdots+\binom{2n+1}{2n-2}(-1)^{n-1}(\cos x)^3(1-\cos^2x)^{n-1}+(-1)^n\cos x(1-\cos^2x)^nx$$

$$\implies\cos(2n+1)x=(\cos x)^{2n+1}\left[1+\binom{2n+1}2+\binom{2n+1}4+\cdots+\binom{2n+1}{2n}\right]+\cdots+\binom{2n+1}{2n}(-1)^n\cos x$$

Now $\displaystyle1+\binom{2n+1}2+\binom{2n+1}4+\cdots+\binom{2n+1}{2n}=\sum_{r=0}^n\binom{2n+1}{2r}=\sum_{r=0}^n\binom{2n+1}{2n+1-2r}=\dfrac{(1+1)^{2n+1}}2=2^{2n}$

$$\implies\cos(2n+1)x=2^{2n}(\cos x)^{2n+1}+\cdots+(-1)^n(2n+1)\cos x$$

Now if $\cos(2n+1)x=\cos A,(2n+1)x=2m\pi\pm A$ where $m$ is any integer

and $x=\dfrac{2m\pi\pm A}{2n+1}$ where $m\equiv0,\pm1,\pm2\cdots,\pm n\pmod{2n+1}$

So, the roots of $$2^{2n}(\cos x)^{2n+1}+\cdots+(-1)^n(2n+1)\cos x-\cos A=0\ \ \ \ (1)$$ are $\cos\dfrac{2m\pi\pm A}{2n+1}$ where $m\equiv0,\pm1,\pm2\cdots,\pm n\pmod{2n+1}$

Using Vieta's formula, $$\cos\dfrac A{2n+1}\prod_{m=1}^n\cos\dfrac{2m\pi\pm A}{2n+1}=\dfrac{\cos A}{2^{2n}}$$

If $A=0,$ $$\prod_{m=1}^n\cos^2\dfrac{2m\pi}{2n+1}=\dfrac1{2^{2n}}$$

Now $\cos\dfrac{2m\pi}{2n+1}=-\cos\left(\pi-\dfrac{2m\pi}{2n+1}\right)=-\cos\dfrac{(2n+1-2m)\pi}{2n+1}$

$$\implies(-1)^n\prod_{m=0}^{2n}\cos\dfrac{m\pi}{2n+1}=\dfrac1{2^{2n}}$$

As $\displaystyle\cos\dfrac{(2n+1-m)\pi}{2n+1}=\cdots=-\cos\dfrac{m\pi}{2m+1}$

$\displaystyle\implies(-1)^n\prod_{m=1}^n\cos^2\dfrac{m\pi}{2n+1}(-1)^n=\dfrac1{2^{2n}}\implies\prod_{m=1}^n\cos^2\dfrac{m\pi}{2n+1}=\dfrac1{2^{2n}}$

Again $0\le\dfrac{m\pi}{2n+1}\le\dfrac\pi2\iff0\le2m\le2n+1\iff0\le m\le n$

$\displaystyle\implies\prod_{m=1}^n\cos\dfrac{m\pi}{2n+1}=\dfrac1{2^n}$ as $\cos\dfrac{m\pi}{2n+1}>0$ as $0<m\le n$

• Please could you tell me how $-\cos\dfrac{(2n+1-2m)\pi}{2n+1} \implies(-1)^n\prod_{m=0}^{2n}\cos\dfrac{m\pi}{2n+1}=\dfrac1{2^{2n}}$ ? I also cannot see how limit $n$ becomes$2n$ Commented Feb 10, 2022 at 18:00

This is definitely not the way to do it. But it sure was a lot of fun to do.

$$\cos {\dfrac{7\pi}{15}} = \cos{\left(\pi - \dfrac{7\pi}{15}\right)} = -\cos{\dfrac{8\pi}{15}}$$

$$\sin{\dfrac{16\pi}{15}} = \sin{\left(\pi + \dfrac{\pi}{15}\right)} = -\sin{\dfrac{\pi}{15}}$$

$$\sin{\dfrac{12\pi}{15}} = \sin{\left(\pi - \dfrac{3\pi}{15}\right)} = \sin{\dfrac{3\pi}{15}}$$

\begin{align} y & = \prod_{r=1}^{7} \cos {\frac{r\pi}{15}}\\ y & = \cos{\frac{\pi}{15}} \cos{\frac{2\pi}{15}} \cos{\frac{3\pi}{15}} \cos{\frac{4\pi}{15}} \cos{\frac{5\pi}{15}}\cos{\frac{6\pi}{15}} \cos{\frac{7\pi}{15}}\\ 2^7y\,\sin{\frac{\pi}{15}} & = 2^6 \left(2\sin{\frac{\pi}{15}} \cos{\frac{\pi}{15}}\right) \cos{\frac{2\pi}{15}} \cos{\frac{3\pi}{15}} \cos{\frac{4\pi}{15}} \cos{\frac{5\pi}{15}}\cos{\frac{6\pi}{15}} \cos{\frac{7\pi}{15}}\\ 2^7y\,\sin{\frac{\pi}{15}} & = 2^5 \left(2\sin{\frac{2\pi}{15}} \cos{\frac{2\pi}{15}}\right) \cos{\frac{3\pi}{15}} \cos{\frac{4\pi}{15}} \cos{\frac{5\pi}{15}}\cos{\frac{6\pi}{15}} \cos{\frac{7\pi}{15}}\\ 2^7y\,\sin{\frac{\pi}{15}} & = 2^4 \left(2\sin{\frac{4\pi}{15}}\right) \cos{\frac{3\pi}{15}} \left( \cos{\frac{4\pi}{15}}\right) \cos{\frac{5\pi}{15}}\cos{\frac{6\pi}{15}} \cos{\frac{7\pi}{15}}\\ 2^7y\,\sin{\frac{\pi}{15}} & = 2^3 \left(2\sin{\frac{8\pi}{15}} \right) \cos{\frac{3\pi}{15}} \cos{\frac{5\pi}{15}}\cos{\frac{6\pi}{15}} \left(-\cos{\frac{8\pi}{15}}\right)\\ 2^7y\,\sin{\frac{\pi}{15}} & = 2^3 \left(-\sin{\frac{16\pi}{15}}\right) \cos{\frac{3\pi}{15}} \cos{\frac{5\pi}{15}}\cos{\frac{6\pi}{15}}\\ 2^7y & = 2^3\cos{\frac{3\pi}{15}} \cos{\frac{5\pi}{15}}\cos{\frac{6\pi}{15}}\\ 2^7\sin{\frac{3\pi}{15}}y & = 2^2\left(2 \sin{\frac{3\pi}{15}}\cos{\frac{3\pi}{15}}\right) \cos{\frac{5\pi}{15}}\cos{\frac{6\pi}{15}}\\ 2^7\sin{\frac{3\pi}{15}}y & = 2\left(2 \sin{\frac{6\pi}{15}}\right) \cos{\frac{5\pi}{15}} \left( \cos{\frac{6\pi}{15}} \right)\\ 2^7\sin{\frac{3\pi}{15}}y & = 2\sin{\frac{12\pi}{15}} \cos{\frac{5\pi}{15}}\\ 2^7\sin{\frac{3\pi}{15}}y & = 2\sin{\frac{3\pi}{15}} \cos{\frac{\pi}{3}}\\ 2^7y & = 1\\ y & = \dfrac{1}{2^7}\\ \end{align}